$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors such that $|\mathbf{a}|=3$, $|\mathbf{b}|=5,|\mathbf{c}|=7$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are perpendicular to the vectors $\mathbf{b}+\mathbf{c}, \mathbf{c}+\mathbf{a}, \mathbf{a}+\mathbf{b}$ respectively, then…

Question

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors such that $|\mathbf{a}|=3$, $|\mathbf{b}|=5,|\mathbf{c}|=7$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are perpendicular to the vectors $\mathbf{b}+\mathbf{c}, \mathbf{c}+\mathbf{a}, \mathbf{a}+\mathbf{b}$ respectively, then $\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})^2-2}=$

Aicharya · Accepted Answer

We have, $ |\mathbf{a}|=3,|\mathbf{b}|=5,|\mathbf{c}|=7 $ $\mathbf{a}$ is perpendicular to $(\mathbf{b}+\mathbf{c})$

and $\mathbf{b}$ is perpendicular to $\mathbf{c}+\mathbf{a}$ So, $\mathbf{b} \cdot(\mathbf{c}+\mathbf{a})=0$

and $\mathbf{c}$ is perpendicular to $(\mathbf{a}+\mathbf{b})$

Adding Eqs. (i), (ii) and (iii), we get

Now, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2$ $ \begin{aligned} & =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\ & =(3)^2+(5)^2+(7)^2+0=83 \end{aligned} $ Now, $\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})^2-2}=\sqrt{83-2}=\sqrt{81}=9$

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three vectors such that $|\mathbf{a}|=3$, $|\mathbf{b}|=5…

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